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VladimirAG [237]
3 years ago
7

Find the area of this trapezoid explain or show your strategy

Mathematics
1 answer:
Dafna11 [192]3 years ago
5 0

sorry if this is wrong but i think the answer is 21

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If the length of a side of an<br> equilateral triangle is x, what is half of the perimeter?
nataly862011 [7]

Answer:

3/2 x

Step-by-step explanation:

length of a side of an  equilateral triangle is x

The perimeter is

P = 3s  where s is the side length

P = 3x

1/2 of the perimeter

1/2 (3x)

3/2 x

4 0
3 years ago
What is the volume of this solid?<br><br> A. 1104<br> B. 132<br> C. 96<br> D. 276
statuscvo [17]

For this case we have that the volume of the figure is composed of the volume of a prism and the volume of a pyramid:

The volume of the prism is given by:

V = A_ {b} * h

Where:

A_ {b}: It is the area of the base

h: It's the height

Substituting:V = 6 * 6 * 6\\V = 216 \ units ^ 3

The volume of the pyramid is given by:

V = \frac {1} {3} * L ^ 2 * h

Where:

L ^ 2:It is the area of the base

h: It's the height

Substituting:

V = \frac {1} {3} * 6 ^ 2 * 5\\V = \frac {1} {3} * 36 * 5\\V = 60units ^ 3

We add and we have:

V = 276 \ units ^ 3

ANswer:

Option D

6 0
3 years ago
HELP ASAP!!! please i know its Friday but i need help with home work
mote1985 [20]
The answer is one solution
8 0
3 years ago
Read 2 more answers
Please help I need it asap!!!!! The equation of line k is y=ax+b. Which could be the equation of line l?
Reika [66]

Answer:

Option D

Step-by-step explanation:

See attached image

3 0
2 years ago
(3x-5)/(x-5)&gt;=0 <br> How do I get the answers for this problem?
Diano4ka-milaya [45]
\frac{3x-5}{x-5} > 0 

First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like: \frac{3x-5}{x-5} = 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x = \frac{5}{3}
Sixth, from the values of x above, we have these 3 intervals to test: 
x < \frac{5}{3}
\frac{5}{3} < x < 5
x > 5
Seventh, pick a test point for each interval. 

1. For the interval x < \frac{5}{3} :
Let's pick x - 0. Then, \frac{3x0-5}{0-5} > 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.

2. For the interval \frac{5}{3} < x < 5:
Let's pick x = 2. Then, \frac{3x2-5}{2-5} > 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.

3. For the interval x > 5:
Let's pick x = 6. Then, \frac{3x6-5}{6-5} > 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x < \frac{5}{3} and x > 5

Answer: x < \frac{5}{3} and x > 5

3 0
3 years ago
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